int_{0}^{2} [x^{2}] , dx | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Consider the given integral $I = \int_{0}^{2} [x^{2}] \, dx$. Since the greatest integer function $[x^2]$ changes its value at $x^2 = 1, 2, 3$,we

Vedclass · Accepted Answer

$ 5-\sqrt{2}-\sqrt{3} $

Vedclass · Answer

(C) Consider the given integral $I = \int_{0}^{2} [x^{2}] \, dx$. Since the greatest integer function $[x^2]$ changes its value at $x^2 = 1, 2, 3$,we split the interval $[0, 2]$ based on $x = 1, \sqrt{2}, \sqrt{3}$. For $x \in [0, 1)$,$x^2 \in [0, 1)$,so $[x^2] = 0$. For $x \in [1, \sqrt{2})$,$x^2 \in [1, 2)$,so $[x^2] = 1$. For $x \in [\sqrt{2}, \sqrt{3})$,$x^2 \in [2, 3)$,so $[x^2] = 2$. For $x \in [\sqrt{3}, 2]$,$x^2 \in [3, 4]$,so $[x^2] = 3$. Thus,$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$. $I = 0 + [x]_{1}^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^{2}$. $I = (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$. $I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$. $I = 5 - \sqrt{2} - \sqrt{3}$.

$ \int_{0}^{2} [x^{2}] \, dx $

Similar Questions

$\int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} = $

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

$3. \int_0^{\frac{1}{2}} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx =$

$\int_{-2}^{2} |[x]| \, dx = $

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module